\begin{frame}
  \frametitle{Differentiation Rules: Quotient Rule}

  Assume that $f$ and $g$ are differentiable at $x$, and define 
  \begin{talign}
    h(x) &= \frac{f(x)}{g(x)}
    &
    \mpause[1]{
    \begin{aligned}
    \Delta h &= h(x+\Delta x) - h(x) \\
    \Delta f &= f(x+\Delta x) - f(x) \\
    \Delta g &= g(x+\Delta x) - g(x)
    \end{aligned}
    }
  \end{talign}
  \pause\pause
  We try to find the derivative of $h$ at $x$:
  \begin{talign}
    \Delta h &= h(x+\Delta x) - h(x)
    \mpause[1]{= \frac{f(x + \Delta x)}{g(x + \Delta x)} - \frac{f(x)}{g(x)}} 
    \mpause[2]{= \frac{f(x) + \Delta f}{g(x) + \Delta g} - \frac{f(x)}{g(x)}} \\
    &\mpause[3]{= \frac{(f(x) + \Delta f)\cdot g(x) - (g(x) + \Delta g)\cdot f(x)}{(g(x) + \Delta g)\cdot g(x)}} 
    \mpause[4]{= \frac{g(x) \Delta f - f(x) \Delta g}{(g(x) + \Delta g)\cdot g(x)}} \\[1ex]
    %
    \mpause[5]{\alert<13->{h'(x)} }&\mpause[6]{= \lim_{\Delta x \to 0} \frac{\Delta h}{\Delta x}} 
    \mpause[7]{= \lim_{\Delta x \to 0} \frac{\frac{g(x) \Delta f - f(x) \Delta g}{(g(x) + \Delta g)\cdot g(x)}}{\Delta x}} 
    \mpause[8]{= \lim_{\Delta x \to 0} \frac{g(x) \frac{\Delta f}{\Delta x} - f(x) \frac{\Delta g}{\Delta x}}{(g(x) + \Delta g)\cdot g(x)}} \\
    &\mpause[9]{= \frac{g(x) \lim_{\Delta x \to 0}\frac{\Delta f}{\Delta x} - f(x) \lim_{\Delta x \to 0}\frac{\Delta g}{\Delta x}}{\lim_{\Delta x \to 0} (g(x) + \Delta g)\cdot g(x)}} 
    \mpause[10]{= \alert<13->{\frac{g(x) f'(x) - f(x) g'(x)}{g(x)^2}} } 
  \end{talign}
\end{frame}